Problem: Find $\lim_{x\to 0}(1+2x)^{^{\frac{1}{\sin(x)}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $e$ (Choice C) C $e^2$ (Choice D) D The limit doesn't exist.
Solution: Substituting $x=0$ into $(1+2x)^{^{\frac{1}{\sin(x)}}}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(1+2x)^{^{\frac{1}{\sin(x)}}}$, we will find $\lim_{x\to 0}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 0}y$. $\ln(y) =\dfrac{\ln(1+2x)}{\sin(x)}$ Substituting $x=0$ into $\dfrac{\ln(1+2x)}{\sin(x)}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 0}\ln(y) \\\\ &=\lim_{x\to 0}\dfrac{\ln(1+2x)}{\sin(x)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(1+2x)]}{\dfrac{d}{dx}[\sin(x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{\left(\dfrac{2}{1+2x}\right)}{\cos(x)} \\\\ &=\dfrac{\left(\dfrac{2}{1}\right)}{1} \gray{\text{Substitution}} \\\\ &=2 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(1+2x)]}{\dfrac{d}{dx}[\sin(x)]}$ actually exists. We found that $\lim_{x\to 0}\ln(y)=2$, which means $\lim_{x\to 0}y=e^2$. [Why?] In conclusion, $\lim_{x\to 0}(1+2x)^{^{\frac{1}{\sin(x)}}}=e^2$.